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$$Let \space I(a) = \int_{0}^{\frac{\pi}{4}} {e^{x} \tan^{a} x} \space . \space Find \space \lim_{a\to\infty}aI(a). $$

Well this is an integral that I'm curently dealing with and so far I've tried to solve it using different approaches. Still, I haven't solved it and what I find always is an undefined answer. Even when I used mathematica it returned an undefined answer. This solution in my opinion is most likely to be the one that works: $$\\$$ Our Integral is $$\int_{0}^{\frac{\pi}{4}}e^x\space \lim_{a\to\infty}[{a\tan^a{x}}]\space dx,$$ letting $$A=\lim_{a\to\infty}{a\tan^a{x}}$$ We know that $$x \in [0,\frac{\pi}{4}]\space \Rightarrow \space \tan{x} \in[0,1] \space \Rightarrow \space \lim_{a\to\infty}a\tan{^ax}= \ \begin{cases} 0 & 0\leq x< \frac{\pi}{4} \\ 1 & x=\frac{\pi}{4}\\ \end{cases} \ $$ Which tells us that $$ A = \ \begin{cases} 0 \times \infty & 0\leq x< \frac{\pi}{4} \\ \infty & x=\frac{\pi}{4}\\ \end{cases} $$ Which are some undefined integrands... . Mathematica says the integral is undefined but the book, Advanced Calculus Explored, says it has an answer. I think this is the right solution and I just need to work on the limit a little bit more, but this is something that I'm stuck in. I appreciate any kind of hints or helps.

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    $\begingroup$ $\lim_{0 \to \infty}$? It is not always valid to say $\lim \int = \int \lim$. $\endgroup$
    – aschepler
    Jun 24 at 12:30
  • $\begingroup$ @Saman Look up the dominated convergence theorem. $\endgroup$
    – Gary
    Jun 24 at 12:50
  • $\begingroup$ Sorry I don't know what it is $\endgroup$
    – Saman
    Jun 24 at 12:56
  • $\begingroup$ Integration by parts gives $I(a) = \exp(\pi/4) - a[I(a-1)+I(a+1)]$, so if the limit exists, it's $\frac{1}{2} \exp(\pi/4)$. Now maybe to find some good squeeze functions? $\endgroup$
    – aschepler
    Jun 24 at 13:03
  • $\begingroup$ @Saman That is why I told you to look it up. $\endgroup$
    – Gary
    Jun 24 at 13:24
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Our problem is a disguised form of the result

$$\tag 1 \lim_{a\to \infty} a\int_0^1 y^a f(y)\,dy=f(1),$$

which holds for any continuous $f$ on $[0,1].$ It has been proved on MSE many times. The proof is quite simple for $f$ continuously differentiable on $[0,1],$ using integration by parts.

To see why the above will help us in the question at hand, let $x=\arctan y.$ The original expression then turns into

$$a\int_0^1 y^a e^{\arctan y}\frac{1}{1+y^2}\,dy.$$

Letting $f(y) = e^{\arctan y}\frac{1}{1+y^2},$ we see that $(1)$ implies our limit is

$$f(1)= e^{\arctan 1}\frac{1}{1+1^2} = \frac{e^{\pi/4}}{2}.$$

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  • $\begingroup$ +1 for the simpler approach. $\endgroup$
    – Paramanand Singh
    Jun 24 at 23:26
  • $\begingroup$ This solution is awesome! I tried to prove it using only integration by parts but I couldn't; by this I mean only with the knowledge of calculus, not real analysis or ... because I haven't studied it yet. Is there any proof without using rigorous analysis? Because I didn't find any of this kind in even the MSE $\endgroup$
    – Saman
    Jun 25 at 13:36
  • $\begingroup$ @Saman The proof of $(1)$ is fairly simple if $f$ is continuously differentiable. Did you try that? Once we have that, there's nothing going on except for the substitution $x=\arctan y.$ $\endgroup$
    – zhw.
    Jun 27 at 14:47
  • $\begingroup$ Thanks. Yes I got it $\endgroup$
    – Saman
    Jun 28 at 11:27
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Rewrite the integral as

$$\int_0^{\pi/4} dx \, \exp{x} \, \exp{\left [a \left (\log{\tan{x}} \right ) \right ]}$$

Sub $x=\pi/4-y$ and use the tangent addition rule and the integral is

$$\exp{\left ( \frac{\pi}{4} \right )} \int_0^{\pi/4} dy \, \exp{(-y)} \, \exp{\left [a \left (\log{\left (\frac{1-\tan{y}}{1+\tan{y}} \right )} \right ) \right ]}$$

Not that the dominant contribution to the integral as $a \to \infty$ is in the interval in a small neighborhood about $y=0$. So expand the integrand about $y=0$; then the interval of integration may be expanded out to $[0,\infty)$ because the other contributions are exponentially subdominant. Accordingly, as $a \to \infty$, the integral behaves as

$$\exp{\left ( \frac{\pi}{4} \right )} \int_0^{\infty} dy \, e^{-2 a y} = \frac1{2 a} \exp{\left ( \frac{\pi}{4} \right )}$$

Accordingly, the sought-after limit is $$ \frac12 \exp{\left ( \frac{\pi}{4} \right )}$$

This has been verified in Mathematica.

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  • $\begingroup$ Thank you sooooo much! It's exactly the answer of the book $\endgroup$
    – Saman
    Jun 24 at 13:12
  • $\begingroup$ @Saman: pro-tip: when trying it out in Mathematica, evaluate the limit numerically; that is, compute a table of values for larger and larger $a$. $\endgroup$
    – Ron Gordon
    Jun 24 at 13:13
  • $\begingroup$ It might clarify things a bit to make explicit that for $y$ near 0, you're expanding $\log \left( \frac{1-\tan y}{1+\tan y} \right) \sim -2y$. $\endgroup$ Jun 24 at 18:56
  • $\begingroup$ This is a special case of Laplace's method (the general method, not the particular set of cases called Laplace's method in the Wikipedia article). $\endgroup$ Jun 24 at 18:57
  • $\begingroup$ @DanielSchepler: I intended to outline just a few steps and give a sense of the solution. I leave minutiae like that to the reader. $\endgroup$
    – Ron Gordon
    Jun 24 at 19:00
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Another approach (that works for this type of limits, with $a$ times an integral of an $a$-th power times something else) is to "absorb" the factor $a$ using integration by parts. Let $g_a(x)=\tan^a x$. Then $a\cdot e^x\tan^a x=f(x)g_a'(x)$, where $f(x)=e^x\tan x\cos^2 x\color{LightGray}{[{}=\ldots]}$. Hence $$aI(a)=\int_0^{\pi/4}f(x)g_a'(x)\,dx=\underbrace{f(\pi/4)g_a(\pi/4)}_{=(1/2)e^{\pi/4}}-\underbrace{f(0)g_a(0)}_{=0}-\int_0^{\pi/4}f'(x)g_a(x)\,dx.$$ The last integral tends to $0$ as $a\to\infty$ (by DCT as noted in comments if you know it, or by bounding the integrand from above: $f'(x)$ by a constant, and $g_a(x)$ by, say, $\tan x\leqslant 4x/\pi$).

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